What is the integral of (x^4)/((x^2+3)^2) ?

The integral of (x^4)/((x^2+3)^2) is x-sqrt(3) 3/2 arctan(x/(sqrt(3)))+(3x)/(2(3+x^2))+C

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integral of (x^4)/((x^2+3)^2)

\int \frac{x ^{4}}{( x ^{2} + 3 ) ^{2}} d x

x - 3 \frac{3}{2} arctan (\frac{x}{3}) + \frac{3 x}{2 ( 3 + x ^{2} )} + C

Solution steps

\int \frac{x ^{4}}{( x ^{2} + 3 ) ^{2}} d x

Apply Integral Substitution

= \int \frac{3 u ^{4}}{( u ^{2} + 1 ) ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{u ^{4}}{( u ^{2} + 1 ) ^{2}} d u

Take the partial fraction of

\frac{u ^{4}}{( u ^{2} + 1 ) ^{2}} : 1 - \frac{2}{u ^{2} + 1} + \frac{1}{( u ^{2} + 1 ) ^{2}}

= 3 \cdot \int 1 - \frac{2}{u ^{2} + 1} + \frac{1}{( u ^{2} + 1 ) ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int 1 d u - \int \frac{2}{u ^{2} + 1} d u + \int \frac{1}{( u ^{2} + 1 ) ^{2}} d u)

\int 1 d u = u

\int \frac{2}{u ^{2} + 1} d u = 2 arctan (u)

\int \frac{1}{( u ^{2} + 1 ) ^{2}} d u = \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}})

= 3 (u - 2 arctan (u) + \frac{1}{2} (arctan (u) + \frac{u}{1 + u ^{2}}))

Substitute back

u = \frac{x}{3}

= 3 \frac{x}{3} - 2 arctan (\frac{x}{3}) + \frac{1}{2} arctan (\frac{x}{3}) + \frac{\frac{x}{3}}{1 + ( \frac{x}{3} ) ^{2}}

Simplify

3 \frac{x}{3} - 2 arctan (\frac{x}{3}) + \frac{1}{2} arctan (\frac{x}{3}) + \frac{\frac{x}{3}}{1 + ( \frac{x}{3} ) ^{2}} : x - 3 \frac{3}{2} arctan (\frac{x}{3}) + \frac{3 x}{2 ( 3 + x ^{2} )}

= x - 3 \frac{3}{2} arctan (\frac{x}{3}) + \frac{3 x}{2 ( 3 + x ^{2} )}

Add a constant to the solution

= x - 3 \frac{3}{2} arctan (\frac{x}{3}) + \frac{3 x}{2 ( 3 + x ^{2} )} + C

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What is the integral of (x^4)/((x^2+3)^2) ?
The integral of (x^4)/((x^2+3)^2) is x-sqrt(3) 3/2 arctan(x/(sqrt(3)))+(3x)/(2(3+x^2))+C